When does a map between CW-complexes give a relative CW-complex?
1
$begingroup$
Let $X$, $Y$ be (absolute) CW-complexes and $varphi:Xto Y$ a continuous map. I would like to know under which assumptions $(Y,varphi(X))$ is a relative CW-complex.
I got interested in this question in a more specific case, i.e. when $Y=X$ and $varphi$ is homotopic to the identity of $X$. However, I don't have many knowledge on algebraic topology to know related results, so every help would be appreciated.
P.S. I already know that this works if $varphi(X)$ is a subcomplex of $Y$.
algebraic-topology cw-complexes
asked Dec 13 '18 at 16:12
Antonio LorenzinAntonio Lorenzin
415
$endgroup$
add a comment |
1
$begingroup$
Let $X$, $Y$ be (absolute) CW-complexes and $varphi:Xto Y$ a continuous map. I would like to know under which assumptions $(Y,varphi(X))$ is a relative CW-complex.
I got interested in this question in a more specific case, i.e. when $Y=X$ and $varphi$ is homotopic to the identity of $X$. However, I don't have many knowledge on algebraic topology to know related results, so every help would be appreciated.
P.S. I already know that this works if $varphi(X)$ is a subcomplex of $Y$.
algebraic-topology cw-complexes
asked Dec 13 '18 at 16:12
Antonio LorenzinAntonio Lorenzin
415
$endgroup$
2
$begingroup$
I believe it is irrelevant whether $X$ is CW-complex or not. The answer only depends on $varphi(X)$, and even if $X$ is a CW-complex, you do not know much about $varphi(X)$. See math.stackexchange.com/q/2505862. By the way, I guess you allow that the cells of $(Y, varphi(X))$ are completely different from those of $Y$?
$endgroup$
– Paul Frost
Dec 13 '18 at 17:34
$begingroup$
Thank you for the comment. I can understand your belief, but I was hoping to find assumption on X rather than on the nature of $varphi$. Also, you guess right, I have no interest in which possible CW structure we give to $(Y, varphi(X))$.
$endgroup$
– Antonio Lorenzin
Dec 13 '18 at 18:22
1
$begingroup$
A minimal requirement is that the inclusion $varphi(X) hookrightarrow Y$ is a cofibration, and I doubt that this can be assured by assumptions on $X$.
$endgroup$
– Paul Frost
Dec 13 '18 at 18:31
3
$begingroup$
Here is an example. Let $X = D^3$ = three dimensional closed ball in $mathbb{R}^3$. It admits an embedding $varphi : D^3 to mathbb{R}^3$ such that the complement of $A = varphi(D^3)$ is not simply connected. An example is the Alexander horned ball (which is enclosed by the Alexander horned sphere en.wikipedia.org/wiki/Alexander_horned_sphere). The quotient $mathbb{R}^3/A$ is not simply connected, therefore $mathbb{R}^3/A$ is not homotopy equivalent to $mathbb{R}^3$ which would be a consequence if $A hookrightarrow mathbb{R}^3$ would be a cofibration.
$endgroup$
– Paul Frost
Dec 13 '18 at 18:45
add a comment |
1
1
1
$begingroup$
Let $X$, $Y$ be (absolute) CW-complexes and $varphi:Xto Y$ a continuous map. I would like to know under which assumptions $(Y,varphi(X))$ is a relative CW-complex.
I got interested in this question in a more specific case, i.e. when $Y=X$ and $varphi$ is homotopic to the identity of $X$. However, I don't have many knowledge on algebraic topology to know related results, so every help would be appreciated.
P.S. I already know that this works if $varphi(X)$ is a subcomplex of $Y$.
algebraic-topology cw-complexes
asked Dec 13 '18 at 16:12
Antonio LorenzinAntonio Lorenzin
415
$endgroup$
Let $X$, $Y$ be (absolute) CW-complexes and $varphi:Xto Y$ a continuous map. I would like to know under which assumptions $(Y,varphi(X))$ is a relative CW-complex.
I got interested in this question in a more specific case, i.e. when $Y=X$ and $varphi$ is homotopic to the identity of $X$. However, I don't have many knowledge on algebraic topology to know related results, so every help would be appreciated.
P.S. I already know that this works if $varphi(X)$ is a subcomplex of $Y$.
algebraic-topology cw-complexes
algebraic-topology cw-complexes
asked Dec 13 '18 at 16:12
Antonio LorenzinAntonio Lorenzin
415
asked Dec 13 '18 at 16:12
Antonio LorenzinAntonio Lorenzin
415
edited Dec 13 '18 at 17:21
Antonio Lorenzin
asked Dec 13 '18 at 16:12
Antonio LorenzinAntonio Lorenzin
415
asked Dec 13 '18 at 16:12
Antonio LorenzinAntonio Lorenzin
415
asked Dec 13 '18 at 16:12
Antonio LorenzinAntonio Lorenzin
415
415
2
$begingroup$
I believe it is irrelevant whether $X$ is CW-complex or not. The answer only depends on $varphi(X)$, and even if $X$ is a CW-complex, you do not know much about $varphi(X)$. See math.stackexchange.com/q/2505862. By the way, I guess you allow that the cells of $(Y, varphi(X))$ are completely different from those of $Y$?
$endgroup$
– Paul Frost
Dec 13 '18 at 17:34
$begingroup$
Thank you for the comment. I can understand your belief, but I was hoping to find assumption on X rather than on the nature of $varphi$. Also, you guess right, I have no interest in which possible CW structure we give to $(Y, varphi(X))$.
$endgroup$
– Antonio Lorenzin
Dec 13 '18 at 18:22
1
$begingroup$
A minimal requirement is that the inclusion $varphi(X) hookrightarrow Y$ is a cofibration, and I doubt that this can be assured by assumptions on $X$.
$endgroup$
– Paul Frost
Dec 13 '18 at 18:31
3
$begingroup$
Here is an example. Let $X = D^3$ = three dimensional closed ball in $mathbb{R}^3$. It admits an embedding $varphi : D^3 to mathbb{R}^3$ such that the complement of $A = varphi(D^3)$ is not simply connected. An example is the Alexander horned ball (which is enclosed by the Alexander horned sphere en.wikipedia.org/wiki/Alexander_horned_sphere). The quotient $mathbb{R}^3/A$ is not simply connected, therefore $mathbb{R}^3/A$ is not homotopy equivalent to $mathbb{R}^3$ which would be a consequence if $A hookrightarrow mathbb{R}^3$ would be a cofibration.
$endgroup$
– Paul Frost
Dec 13 '18 at 18:45
add a comment |
2
$begingroup$
I believe it is irrelevant whether $X$ is CW-complex or not. The answer only depends on $varphi(X)$, and even if $X$ is a CW-complex, you do not know much about $varphi(X)$. See math.stackexchange.com/q/2505862. By the way, I guess you allow that the cells of $(Y, varphi(X))$ are completely different from those of $Y$?
$endgroup$
– Paul Frost
Dec 13 '18 at 17:34
$begingroup$
Thank you for the comment. I can understand your belief, but I was hoping to find assumption on X rather than on the nature of $varphi$. Also, you guess right, I have no interest in which possible CW structure we give to $(Y, varphi(X))$.
$endgroup$
– Antonio Lorenzin
Dec 13 '18 at 18:22
1
$begingroup$
A minimal requirement is that the inclusion $varphi(X) hookrightarrow Y$ is a cofibration, and I doubt that this can be assured by assumptions on $X$.
$endgroup$
– Paul Frost
Dec 13 '18 at 18:31
3
$begingroup$
Here is an example. Let $X = D^3$ = three dimensional closed ball in $mathbb{R}^3$. It admits an embedding $varphi : D^3 to mathbb{R}^3$ such that the complement of $A = varphi(D^3)$ is not simply connected. An example is the Alexander horned ball (which is enclosed by the Alexander horned sphere en.wikipedia.org/wiki/Alexander_horned_sphere). The quotient $mathbb{R}^3/A$ is not simply connected, therefore $mathbb{R}^3/A$ is not homotopy equivalent to $mathbb{R}^3$ which would be a consequence if $A hookrightarrow mathbb{R}^3$ would be a cofibration.
$endgroup$
– Paul Frost
Dec 13 '18 at 18:45
2
2
$begingroup$
I believe it is irrelevant whether $X$ is CW-complex or not. The answer only depends on $varphi(X)$, and even if $X$ is a CW-complex, you do not know much about $varphi(X)$. See math.stackexchange.com/q/2505862. By the way, I guess you allow that the cells of $(Y, varphi(X))$ are completely different from those of $Y$?
$endgroup$
– Paul Frost
Dec 13 '18 at 17:34
$begingroup$
I believe it is irrelevant whether $X$ is CW-complex or not. The answer only depends on $varphi(X)$, and even if $X$ is a CW-complex, you do not know much about $varphi(X)$. See math.stackexchange.com/q/2505862. By the way, I guess you allow that the cells of $(Y, varphi(X))$ are completely different from those of $Y$?
$endgroup$
– Paul Frost
Dec 13 '18 at 17:34
$begingroup$
Thank you for the comment. I can understand your belief, but I was hoping to find assumption on X rather than on the nature of $varphi$. Also, you guess right, I have no interest in which possible CW structure we give to $(Y, varphi(X))$.
$endgroup$
– Antonio Lorenzin
Dec 13 '18 at 18:22
$begingroup$
Thank you for the comment. I can understand your belief, but I was hoping to find assumption on X rather than on the nature of $varphi$. Also, you guess right, I have no interest in which possible CW structure we give to $(Y, varphi(X))$.
$endgroup$
– Antonio Lorenzin
Dec 13 '18 at 18:22
1
1
$begingroup$
A minimal requirement is that the inclusion $varphi(X) hookrightarrow Y$ is a cofibration, and I doubt that this can be assured by assumptions on $X$.
$endgroup$
– Paul Frost
Dec 13 '18 at 18:31
$begingroup$
A minimal requirement is that the inclusion $varphi(X) hookrightarrow Y$ is a cofibration, and I doubt that this can be assured by assumptions on $X$.
$endgroup$
– Paul Frost
Dec 13 '18 at 18:31
3
3
$begingroup$
Here is an example. Let $X = D^3$ = three dimensional closed ball in $mathbb{R}^3$. It admits an embedding $varphi : D^3 to mathbb{R}^3$ such that the complement of $A = varphi(D^3)$ is not simply connected. An example is the Alexander horned ball (which is enclosed by the Alexander horned sphere en.wikipedia.org/wiki/Alexander_horned_sphere). The quotient $mathbb{R}^3/A$ is not simply connected, therefore $mathbb{R}^3/A$ is not homotopy equivalent to $mathbb{R}^3$ which would be a consequence if $A hookrightarrow mathbb{R}^3$ would be a cofibration.
$endgroup$
– Paul Frost
Dec 13 '18 at 18:45
$begingroup$
Here is an example. Let $X = D^3$ = three dimensional closed ball in $mathbb{R}^3$. It admits an embedding $varphi : D^3 to mathbb{R}^3$ such that the complement of $A = varphi(D^3)$ is not simply connected. An example is the Alexander horned ball (which is enclosed by the Alexander horned sphere en.wikipedia.org/wiki/Alexander_horned_sphere). The quotient $mathbb{R}^3/A$ is not simply connected, therefore $mathbb{R}^3/A$ is not homotopy equivalent to $mathbb{R}^3$ which would be a consequence if $A hookrightarrow mathbb{R}^3$ would be a cofibration.
$endgroup$
– Paul Frost
Dec 13 '18 at 18:45
add a comment |
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2
$begingroup$
I believe it is irrelevant whether $X$ is CW-complex or not. The answer only depends on $varphi(X)$, and even if $X$ is a CW-complex, you do not know much about $varphi(X)$. See math.stackexchange.com/q/2505862. By the way, I guess you allow that the cells of $(Y, varphi(X))$ are completely different from those of $Y$?
$endgroup$
– Paul Frost
Dec 13 '18 at 17:34
$begingroup$
Thank you for the comment. I can understand your belief, but I was hoping to find assumption on X rather than on the nature of $varphi$. Also, you guess right, I have no interest in which possible CW structure we give to $(Y, varphi(X))$.
$endgroup$
– Antonio Lorenzin
Dec 13 '18 at 18:22
1
$begingroup$
A minimal requirement is that the inclusion $varphi(X) hookrightarrow Y$ is a cofibration, and I doubt that this can be assured by assumptions on $X$.
$endgroup$
– Paul Frost
Dec 13 '18 at 18:31
3
$begingroup$
Here is an example. Let $X = D^3$ = three dimensional closed ball in $mathbb{R}^3$. It admits an embedding $varphi : D^3 to mathbb{R}^3$ such that the complement of $A = varphi(D^3)$ is not simply connected. An example is the Alexander horned ball (which is enclosed by the Alexander horned sphere en.wikipedia.org/wiki/Alexander_horned_sphere). The quotient $mathbb{R}^3/A$ is not simply connected, therefore $mathbb{R}^3/A$ is not homotopy equivalent to $mathbb{R}^3$ which would be a consequence if $A hookrightarrow mathbb{R}^3$ would be a cofibration.
$endgroup$
– Paul Frost
Dec 13 '18 at 18:45